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  3. to prepare 0.250 l of 0.100 m aqueous nacl (58.4 g

To prepare 0.250 L of 0.100 M aqueous NaCl (58.4 g/mol), one may

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Chemistry

To prepare 0.250 L of 0.100 M aqueous NaCl (58.4g/mol), one may

Answer & Explanation Solved by verified expert
4.3 Ratings (633 Votes)

WE KNOW THAT

Molarity = Number of mole of solute/Volume of solution.

        M= Number of mole of NaCl / Volume of solution.

      0.100 = Number of mole of NaCl /0.250L.

        Than

              Number of mole of NaCl = 0.100*0.250

                                                             = 0.0250.

      SO Mass of NaCl = mole* molecular weight

                                        0.0250*58.4 =1.46g
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